Problem: $f(x,y) = \cos(y) - y^3$ What is $\dfrac{\partial f}{\partial x}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\sin(y) - 2y^2$ (Choice B) B $-\sin(x) - 3x^2$ (Choice C) C $0$ (Choice D) D $\cos(y) - y^2$
Solution: We want to find $\dfrac{\partial f}{\partial x}$, which is the partial derivative of $f$ with respect to $x$. When we take a partial derivative with respect to $x$, we treat $y$ as if it were a constant. Let's break $f(x, y)$ down term by term. $\begin{aligned} &\dfrac{\partial}{\partial x} \left[ \cos(y) \right] = 0 \\ \\ &\dfrac{\partial}{\partial x} \left[ -y^3 \right] = 0 \end{aligned}$ Adding the terms back together, we get the partial derivative. In conclusion: $\dfrac{\partial f}{\partial x} = 0 + 0 = 0$